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K⁺ + e⁻ → K . 1 mol K (39 g) requires 1F. Moles = (6.75/39) = 0.1731 mol , Charge = 0.1731 × 96500 = 16700 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
K⁺ + e⁻ → K . 1 mol K (39 g) requires 1F. Moles = (6.75/39) = 0.1731 mol , Charge = 0.1731 × 96500 = 16700 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Ba²⁺ + 2e⁻ → Ba . 1 mol Ba (137 g) requires 2F. Moles = (0.965/137) = 0.00704 mol , Charge = 0.00704 × 2 × 96500 = 1358.24 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Ba²⁺ + 2e⁻ → Ba . 1 mol Ba (137 g) requires 2F. Moles = (0.965/137) = 0.00704 mol , Charge = 0.00704 × 2 × 96500 = 1358.24 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
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